## Thursday, December 26, 2013

### More Laurent Series Examples

No, we’re not yet finished with these exotic mathematical denizens! We want to give yet more examples, and especially showing different ways to obtain the Laurent equivalent of different functions.

Ex. (1):   exp(2z)/ (z - 1)3  about  z 0  =   1

Let u = (z – 1)  so that z = (u -1)

Then:

exp ( 2z) / (z - 1)3     =  exp(2 + 2u) /  u 3  =   [( e) 2  /  u 3   ] /  e 2u

=   e 2  /  u 3      [ 1  + 2u   + (2u) 2 / 2!   +   (2u) 3 / 3!   +

(2u) 4 / 4! +     …….

=  e 2  /  (z - 1)3     +   2 e 2  /  (z - 1)2      +   2 e 2  /  (z – 1)  +

4 e 2  / 3  +  2 e 2  / 3 (z – 1)  +   …….

Note that z is a pole of order 3 or a “triple pole” (Why?)    Note also the series converges for all values of z  ¹  1

Ex. (2):   f(z) =  z – sin z /  z 3

We’ve already seen the series for sin z:

sin(z) = z -  z3 / 3!  +  z5 / 5!  …..  -  (-1) n- 1 z 2n -1/  (2n -1)!  +

Then:
z – sin z /  z 3     =  1/  z 3    [z -  (z -  z3 / 3!  +  z5 / 5!  …..  ) ]

=  1/  z 3    [z3 / 3!   -  z5 / 5!    +   z 7 / 7!  -    ……….

=   1/3!  -  z 2 / 5!     +   z 4 / 7!  -    z 6 / 9!   + ……..

Ex. (3)
Find a Laurent series for f(z) =  sin (4z) / z 4

Using the sine series we can write:

sin (4z) / z 4     =  1/   z 4   [4z -    4z3 / 3!  +  4 z5 / 5!  +  ……..]

=     4 /   z 3    -    43 / 3!  z  +   45  z  / 5!       +   …..

= 4 z  -3    -  43  z -1 /  3!   +   45  z  / 5!        +   …..

Which on inspection is found to have the form:

å¥ n = 0    (-1) n   4 2n+ 1  z 2n -3  /   (2n +  1)!
Check term number 3 (n = 2):

4 2n+ 1     = 4 2(2)+ 1      =      45

z 2n -3  =  z 2(2) - 3      = z 1=  z

(2n +  1)!   =  (2(2) +  1)!  = 5!

Then:  we get  45  z  / 5!

A removable singularity occurs at z = 0 so we expect the region of convergence to be:

÷ z  ÷      >   0

Problems for Math Mavens:

Let f(z) = exp (-1/z 2) /  z5

a) Show the Laurent series can be written:

å¥ n = 0    (-1) n   /   n !  z 2n + 5

b)Specify where the singularity would occur and the precise region of convergence

Challenge Problem: For über Math Mavens only –

Let f(z) = 7 z 2  + 9 z  - 18 /  z 3 -  9z

Find Laurent series for the convergence regions:

a)  0  < ÷ z ÷   <    3  and
b)    ÷ z  ÷      >   3

(Hint: Approach the function f(z) by resort to partial fraction decomposition)